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MHT CET 2026 (18 Apr, Shift 1) · Density, Compressibility & Pascal's Law

Density, Compressibility & Pascal's Law question from MHT CET 2026 (18 Apr, Shift 1)

Q3MHT CET 2026 · 18 APR · SHIFT 1Easy

Density, Compressibility & Pascal's Law

A piston of cross-sectional area $2.5 \times 10^{-2}$ m $^{2}$ is used in a hydraulic lift to exert a force of $250$ N on water. The cross-sectional area of the other piston which supports a car of mass $3000$ kg is ( $g = 9.8$ m/s $^{2}$ )

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Answer: (b)

Pascal's law: $\dfrac{F_1}{A_1} = \dfrac{F_2}{A_2} \Rightarrow A_2 = \dfrac{F_2 A_1}{F_1}$ .
$F_2 = 3000\times9.8 = 29400\text{ N}$
$A_2 = \dfrac{29400\times2.5\times10^{-2}}{250} = 2.94\text{ m}^2$
So the answer is (b).

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